【poj】3187全排列

E - E Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status

Description

FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single number is left. For example, one instance of the game (when N=4) might go like this: 3 1 2 4 4 3 6 7 9 16 Behind FJ's back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from only the final total and the number N. Unfortunately, the game is a bit above FJ's mental arithmetic capabilities. Write a program to help FJ play the game and keep up with the cows.

Input

Line 1: Two space-separated integers: N and the final sum.

Output

Line 1: An ordering of the integers 1..N that leads to the given sum. If there are multiple solutions, choose the one that is lexicographically least, i.e., that puts smaller numbers first.

Sample Input

4 16

Sample Output

3 1 2 4

Hint

Explanation of the sample: There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest. 真的很渣呀，对STL了解的少之又少。。。发现好多题都用到STL里的东西    又长知识了：next_permutation()用于全排列http://blog.csdn.net/wconvey/article/details/7459131 coed： #include<cstdio> #include<algorithm> using namespace std; int a[100],b[100]; int main() { int n,sum; while(~scanf("%d%d",&n,&sum)){ for(int i=1;i<=n;i++) a[i]=i; while(next_permutation(a+1,a+n+1)){ for(int i=1;i<=n;i++) b[i]=a[i]; int k=n; while(k--){ for(int i=1;i<=k;i++) b[i]+=b[i+1]; } if(b[1]==sum) break; } for(int i=1;i<=n;i++) printf("%d ",a[i]); printf("\n"); } return 0; }